www.studyguide.pk
UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
9709/42
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.
Mark schemes must be read in conjunction with the question papers and the report on the examination.
• CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the October/November 2009 question papers for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.xtremepapers.net

Page 2
Mark Scheme Notes
Mark Scheme: Teachers’ version
GCE A/AS LEVEL – October/November 2009
Marks are of the following three types: www.studyguide.pk
Syllabus
9709
Paper
42
M Method mark, awarded for a valid method applied to the problem. Method marks are not lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, e.g. by substituting the relevant quantities into the formula. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer.
A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated method mark is earned (or implied).
B Mark for a correct result or statement independent of method marks.
• When a part of a question has two or more “method” steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme.
When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given.
• The symbol √ implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working.
• Note: B2 or A2 means that the candidate can earn 2 or 0.
B2/1/0 means that the candidate can earn anything from 0 to 2.
The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored.
• Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise.
• For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.
© UCLES 2009 www.xtremepapers.net

Page 3 Mark Scheme: Teachers’ version
GCE A/AS LEVEL – October/November 2009 www.studyguide.pk
Syllabus
9709
The following abbreviations may be used in a mark scheme or used on the scripts:
Any Equivalent Form (of answer is equally acceptable) AEF
Paper
42
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear)
CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed)
CWO Correct Working Only – often written by a ‘fortuitous’ answer
ISW Ignore Subsequent Working
MR
PA
Misread
Premature Approximation (resulting in basically correct work that is insufficiently accurate)
SOS
SR
See Other Solution (the candidate makes a better attempt at the same question)
Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance)
Penalties
MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through √”
PA –1 This is deducted from A or B marks in the case of premature approximation. The
PA –1 penalty is usually discussed at the meeting. marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR –2 penalty may be applied in particular cases if agreed at the coordination meeting.
© UCLES 2009 www.xtremepapers.net

Page 4 Mark Scheme: Teachers’ version
GCE A/AS LEVEL – October/November 2009 www.studyguide.pk
Syllabus
9709
Paper
42
1 (i)
[P = Wsin40 o
] M1
For resolving forces parallel to the plane or for a correct triangle of forces or for resolving horizontally and vertically
P = 7.71 A1 2
(ii) [Pcos40 o
= Wsin40 o
] M1
A1 2 P = 10.1
2 (i) Loss in PE is 2.7 × 10
6
J
(ii) WD is 2.1 × 10
6
J
B1 1
B1ft 1 ft incorrect loss in PE
(iii) KE change = ½ 15000(16
2
– 14
2
) B1
[WD = ½ 15000(16
2
WD is 4.45 × 10
6
– 14
2
) + 1600 × 2500] M1
WD by DF = Gain in KE + WD by resistance
SR for candidates who use Newton’s
Law method instead of energy (max 1/3) a = (16
2
– 14
2
)/(2 × 2500) = 0.012
DF = 1600 + 15000 × 0.012 = 1780
WD = 1780 × 2500 = 4.45 × 10
6
B1
M1 For using DF = R at max. speed
M1 For using DF = P/v
3 (i) [DF = 600 at max speed]
[DF = 24000/v]
Speed cannot exceed 40 ms
–1
(ii) DF – R = ma]
24000/15 – 600 = 1250a
Acceleration is 0.8 ms
–2
M1
A1
For using Newton’s second law
A1 3
Mass is 0.125 kg
M1 For resolving forces normal to the plane
A1 2
– 0.125 × 10 × 0.28 – 0.4 = 0.125a a = –6 deceleration is 6 ms
–2
M1 For using Newton’s second law
A1ft ft incorrect mass
A1 3
(iii) M1 For comparing magnitudes of µR (0.4) and mg sinα (0.35)
A1 2 µR > mg sinα particle remains at rest
© UCLES 2009 www.xtremepapers.net

Page 5 Mark Scheme: Teachers’ version
GCE A/AS LEVEL – October/November 2009 www.studyguide.pk
Syllabus
9709
Paper
42
5 (i)
12 + 15sin30 o
F = 15cos30 o
[µ = 15cos30
= R
M1
A1
For resolving forces vertically
B1 o
/(12 + 15sin30 o
] M1 For using µ = F/R
Coefficient is 0.666
(ii) F = 0.666(12 – 15sin30 o
) B1
15cos30 o
– F = 1.2a
Acceleration is 8.33 ms
–2
M1
A1
For using Newton’s second law
A1 4
6 (i) For applying Newton’s second law to A or to B or for using
M1 (M + m)a = (M – m)g
A1 T – 0.3g = 0.3a and 0.7g – T = 0.7a or
(0.7 + 0.3)a = (0.7 – 0.3)g
Acceleration is 4 ms
–2 s
1
= 1.6
2
/(2 × 4)
Height is 0.448 m
A1 3
B1ft
M1 ft acceleration
For using 0
+ s
2
2
= 1.6
2
– 2gs
2
= 0.32 + 0.128
Time taken is 0.56 s
. . . . . . . . . . . . . . . .
(Alternative for part (iii)) t
1
+ t
2
= (s
1
+ s
2
)/0.8
Time taken is 0.56 s
. . . . . . . . . . . . . . . .
(Alternatively for parts ii and iii using v–t
graph)
B1ft ft acceleration
(can be scored in (ii) )
M1 For using 0 = 1.6 – gt
2
A1 3 From t
1
+ t
2
= 0.4 + 0.16
. . . . . . . . . . . . . . . . .
For observing that the average speed is the same for each of the two phases and
M1
A1 equal to
(0 + 1.6)/2 ms
–1
A1 3
[Similarly for finding s found before ans (ii) ]
1
+ s
2
if ans (iii) is
. . . . . . . . . . . . . . . . . t
1
= 1.6/4 and t
2
= 1.6/10
Time taken is 0.56s s
1
= 0.4 × 1.6/2 or s
2
= 0.16 × 1.6/2 or
s
1
+ s
2
= (0.4 + 0.16) × 1.6/2
Height is 0.448m
M1 Use of gradient to find t
1
or t
2
A1
A1
M1
For use of area to find s
1 or s
2 or s
1
+ s
2
A1
A1 6
© UCLES 2009 www.xtremepapers.net

Page 6 Mark Scheme: Teachers’ version
GCE A/AS LEVEL – October/November 2009 www.studyguide.pk
Syllabus
9709
Paper
42
7 (i) v = 1.2t – 0.012t
2
M1 For using v(t) = s &
(t)
A1
For using a(t) =
& v
(t) and evaluating
[a(50) = 1.2 – 0.024 × 50] M1 a(50)
A1 AG
B1 5 a = 0
V = 30

1


= 0.6 × 50
1000
50 +
+ t
2 s
2
2
=
– 0.004 × 50
3
(= 1000)
27 .
5



[1000 + 30t
2 t
2
= 150 t = 200
= 27.5(50 + t
2
)]
. . . . . . . . . . . . . . . .
(Alternative for part (ii)) s
1
= 0.6 × 50
2
– 0.004 × 50
3
(= 1000)
[(1000 + s
2
)/t = 27.5]
(1000 + 30(t – 50))/t = 27.5
[27.5t = 1000 + 30(t – 50)] t = 200
B1
For using ‘average speed = total distance
M1 /total time’
For substituting s
2
M1
A1 to solve for t
2
= Vt
2
and attempting
A1 5 ft 50 + t
2
(requires both M marks)
. . . . . . . . . . . . . . . . .
B1
For using ‘average speed = total distance
M1 /total time’ with t
2
= t – 50
A1ft (ft V and s
1
)
M1 For attempting to solve for t
A1 5
© UCLES 2009 www.xtremepapers.net